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== 10.1#h(0.6em)Polar Coordinates


    The Cartesian coordinate system is also called the #strong[rectangular coordinate system], because it describes a location in the plane as the vertex of a rectangle. To construct a rectangular coordinate system, we begin with two perpendicular axes that intersect at the origin. The #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-coordinates of a point indicate the length and width of a rectangle with one vertex at the origin. The point #math.equation(block: false, alt: "P open parenthesis x , y close parenthesis")[$P ( x , y )$] sits at the opposite vertex of the rectangle, as shown in the figure below.


    In Chapter 9 we used vectors to specify a location by giving a distance and a direction. For example, we might say that the airport is located 5 miles southwest of the city center. This method of designating locations is so useful that we will construct a new coordinate system using the same tools: distance and direction.


    #figure(figph[Cartesian], alt: "Cartesian", caption: none)


    #figure(figph[polar], alt: "polar", caption: none)


    In the #strong[polar coordinate system] we start with the origin or #strong[pole], and a single ray from the pole, called the #strong[polar axis]. We describe the location of a point #math.equation(block: false, alt: "P")[$P$] in the plane by measuring the distance, #math.equation(block: false, alt: "vertical bar r vertical bar")[$| r |$], from #math.equation(block: false, alt: "P")[$P$] to the pole, and the angle, #math.equation(block: false, alt: "θ")[$θ$], that #math.equation(block: false, alt: "vector O P")[$arrow(O P)$] makes with the polar axis (measured counterclockwise). The components of the ordered pair #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] are called the #strong[polar coordinates] of the point #math.equation(block: false, alt: "P")[$P$].


    The two measurements this paragraph describes, both on sliders. The blue ray leaves the pole at angle θ from the polar axis, the marked point sits distance r along it, and the red arc shows the angle actually swept. The dashed grey circle is every point at that same distance r — the whole set the first measurement alone leaves you with, which is why the angle is needed too. θ steps by π/12, so the quarter-turns are among the settings: at θ = π/2 the point stands straight up the vertical axis, at θ = π it lies straight back along the negative polar axis, and at θ = 2π it has returned to where it began — the same point as θ = 0, the first hint that polar coordinates are not unique. The dashed segment joining the point to the horizontal axis marks off the rectangular coordinates: x = r cos θ across, y = r sin θ up or down, negative below the axis.{"functions":\[{"color":"\#6a6a6a","expression":"r\*cos(t)","expression2":"r\*sin(t)","id":"distance-circle","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"t\*r\*cos(th)","expression2":"t\*r\*sin(th)","id":"ray","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"0.8\*cos(th\*t)","expression2":"0.8\*sin(th\*t)","id":"angle-arc","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"r\*cos(th)","expression2":"t\*r\*sin(th)","id":"xy-drop","inequality":"lt","lineDash":\[4,4\],"lineWidth":1.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#b8442a","expression":"r\*cos(th) + 0.12\*cos(t)","expression2":"r\*sin(th) + 0.12\*sin(t)","id":"the-point","inequality":"lt","lineDash":\[\],"lineWidth":3,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-r","label":"Distance r from the pole","max":4,"min":0.5,"name":"r","step":0.5,"value":3},{"id":"p-th","label":"Angle \\u03b8 from the polar axis, in radians (steps of \\u03c0/12)","max":6.2831853072,"min":0,"name":"th","step":0.2617993878,"value":1.0471975512}\],"title":"Polar coordinates: the point (r, \\u03b8)","version":1,"viewport":{"centerX":0,"centerY":0,"scale":46}}#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Polar Coordinates]


      The #strong[polar coordinates] of a point #math.equation(block: false, alt: "P")[$P$] in the plane are #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$], where


      - #math.equation(block: false, alt: "vertical bar r vertical bar")[$| r |$]  is the #strong[distance] from  #math.equation(block: false, alt: "P")[$P$]  to the pole,
- #math.equation(block: false, alt: "θ")[$θ$]   is the #strong[angle] measured counterclockwise from the polar axis to the ray through #math.equation(block: false, alt: "P")[$P$] from the pole.
]


    In our work with polar coordinates, we will always use radians for the angle #math.equation(block: false, alt: "θ")[$θ$]. For example, the point #math.equation(block: false, alt: "P open parenthesis 2 , the fraction π over 2 close parenthesis")[$P ( 2 , display(frac(π, 2)) )$] is located 2 units from the pole at an angle of #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$] radians, and #math.equation(block: false, alt: "Q open parenthesis 3 , 4 close parenthesis")[$Q ( 3 , 4 )$] is located 3 units from the pole at an angle of 4 radians. The graphs of #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$] are shown at right.


    #figure(figph[points in polar coordinates], alt: "points in polar coordinates", caption: none)


    A piece of graph paper in rectangular coordinates consists of a grid of horizontal and vertical lines. These are the lines #math.equation(block: false, alt: "x equals k")[$x = k$] and #math.equation(block: false, alt: "y equals k")[$y = k$], for evenly spaced values of #math.equation(block: false, alt: "k")[$k$]. Each vertical grid line consists of points that have the same #math.equation(block: false, alt: "x")[$x$]-coordinate, and each horizontal grid line is made up of points that have the same #math.equation(block: false, alt: "y")[$y$]-coordinate. The grid lines are bench marks to help us locate points with specific rectangular coordinates.


    #figure(figph[rectangular and polar grids], alt: "rectangular and polar grids", caption: none)


    A piece of graph paper in polar coordinates consists of a grid of concentric circles and radial lines, as shown below.  Each circle consists of points with the same #math.equation(block: false, alt: "r")[$r$]-coordinate; they are all the same distance from the pole. All points with the same #math.equation(block: false, alt: "θ")[$θ$]-coordinate lie on one of the radial lines. This grid helps us locate points with specific polar coordinates.


    
      
=== Plotting Points


      You are used to thinking in rectangular coordinates: to locate a point, we move so many units left or right, and so many units up or down. When working in polar coordinates, we want to "think radially:" how far to move from the pole, and in which direction.


      #examplebox("Example 1")[][
Plot the points whose polar coordinates are given: #math.equation(block: false, alt: "A open parenthesis 1 , the fraction π over 2 close parenthesis")[$A ( 1 , display(frac(π, 2)) )$] and #math.equation(block: false, alt: "B open parenthesis 2 , the fraction 7 π over 4 close parenthesis")[$B ( 2 , display(frac(7 π, 4)) )$].


        
        #solutionbox[
To plot point #math.equation(block: false, alt: "A")[$A$], we move 1 unit away from the pole in the direction #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], as shown at right. To plot point #math.equation(block: false, alt: "B")[$B$], we move 2 units away from the pole in the direction #math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$].


          #figure(figph[polar points], alt: "polar points", caption: none)
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Note that even though the independent or input variable in polar coordinates is nearly always #math.equation(block: false, alt: "θ")[$θ$], it is customary to list the dependent or output variable first: #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$].
]


      
        Give the polar coordinates of the points #math.equation(block: false, alt: "S")[$S$] and #math.equation(block: false, alt: "T")[$T$] shown at right, with #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].


        #figure(figph[polar points], alt: "polar points", caption: none)


        
          #math.equation(block: true, alt: "S open parenthesis 2 , π close parenthesis")[$S ( 2 , π )$], #math.equation(block: true, alt: "T open parenthesis 3 , the fraction 2 π over 3 close parenthesis")[$T ( 3 , display(frac(2 π, 3)) )$]


        
      
      Each point has infinitely many polar coordinates, because we can add multiples of #math.equation(block: false, alt: "2 π")[$2 π$] to the value of #math.equation(block: false, alt: "θ")[$θ$]. For example, the point with polar coordinates #math.equation(block: false, alt: "open parenthesis 1 , the fraction π over 2 close parenthesis")[$( 1 , display(frac(π, 2)) )$] also has polar coordinates #math.equation(block: false, alt: "open parenthesis 1 , the fraction 5 π over 2 close parenthesis")[$( 1 , display(frac(5 π, 2)) )$] and #math.equation(block: false, alt: "open parenthesis 1 , the fraction minus 3 π over 2 close parenthesis")[$( 1 , display(frac(− 3 π, 2)) )$].


      If we allow negative values of #math.equation(block: false, alt: "r")[$r$], there are even more ways to write the coordinates of a point. Each angle designates a line through the pole, and each line has both a postive and a negative direction.


      On the line at #math.equation(block: false, alt: "θ equals the fraction π over 4")[$θ = display(frac(π, 4))$], for instance, the positive direction lies in the first quadrant, so that the point #math.equation(block: false, alt: "open parenthesis 2 , the fraction π over 4 close parenthesis")[$( 2 , display(frac(π, 4)) )$] is designated by #math.equation(block: false, alt: "P")[$P$] in the figure at right. To plot the point #math.equation(block: false, alt: "open parenthesis minus 2 , the fraction π over 4 close parenthesis")[$( − 2 , display(frac(π, 4)) )$], we move in the opposite direction from the pole, arriving at point #math.equation(block: false, alt: "Q")[$Q$]. Note that the point #math.equation(block: false, alt: "Q")[$Q$] can also be designated by the coordinates #math.equation(block: false, alt: "open parenthesis 2 , the fraction 5 π over 4 close parenthesis")[$( 2 , display(frac(5 π, 4)) )$].


      #figure(figph[point on polar grid], alt: "point on polar grid", caption: none)


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Non-Uniqueness of Polar Coordinates]


        + Any point with polar coordinates #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] also has coordinates #math.equation(block: false, alt: "open parenthesis r , θ plus 2 k π close parenthesis")[$( r , θ + 2 k π )$], where #math.equation(block: false, alt: "k")[$k$] is an integer.
+ The point #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] can also be designated by #math.equation(block: false, alt: "open parenthesis minus r , θ plus π close parenthesis")[$( − r , θ + π )$].
+ The pole has coordinates #math.equation(block: false, alt: "open parenthesis 0 , θ close parenthesis")[$( 0 , θ )$], for any value of #math.equation(block: false, alt: "θ")[$θ$].
]


      #examplebox("Example 2")[][
Give polar coordinates with negative #math.equation(block: false, alt: "r")[$r$]-values for the points in the previous Exercise.


        
        #solutionbox[
The point #math.equation(block: false, alt: "S open parenthesis 2 , π close parenthesis")[$S ( 2 , π )$] is also #math.equation(block: false, alt: "open parenthesis minus 2 , π close parenthesis")[$( − 2 , π )$] or #math.equation(block: false, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$]. The point #math.equation(block: false, alt: "T open parenthesis 3 , the fraction 2 π over 3 close parenthesis")[$T ( 3 , display(frac(2 π, 3)) )$] can also be designated by #math.equation(block: false, alt: "open parenthesis minus 3 , the fraction 2 π over 3 plus π close parenthesis")[$( − 3 , display(frac(2 π, 3)) + π )$], or #math.equation(block: false, alt: "open parenthesis minus 3 , the fraction 5 π over 3 close parenthesis")[$( − 3 , display(frac(5 π, 3)) )$].
]
]


      
        Give polar coordinates with positive #math.equation(block: false, alt: "r")[$r$]-values and #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$] for each point given in polar coordinates.


        + #math.equation(block: false, alt: "open parenthesis minus 2.5 , the fraction π over 6 close parenthesis")[$( − 2.5 , display(frac(π, 6)) )$]
+ #math.equation(block: false, alt: "open parenthesis minus 0.7 , 5 close parenthesis")[$( − 0.7 , 5 )$]


        
          + #math.equation(block: false, alt: "open parenthesis 2.5 , the fraction 7 π over 6 close parenthesis")[$( 2.5 , display(frac(7 π, 6)) )$]
+ #math.equation(block: false, alt: "open parenthesis 0.7 , 5 minus π close parenthesis")[$( 0.7 , 5 − π )$]


        
      
    
    
      
=== Regions in the Plane


      In Cartesian coordinates, we use equations and inequalities to describe regions in the plane. For example, the shaded region in figure (a) is the graph of #math.equation(block: false, alt: "y less than or equal to x plus 2")[$y ≤ x + 2$].  The pair of inequalities #math.equation(block: false, alt: "x greater than or equal to minus 2 , 1 less than y less than 3")[$x ≥ − 2 , " " 1 < y < 3$]   describes the region in figure (b). This region is particularly simple, in that its boundaries are portions of the coordinate grid lines #math.equation(block: false, alt: "x equals k")[$x = k$] and #math.equation(block: false, alt: "y equals k")[$y = k$], where #math.equation(block: false, alt: "k")[$k$] is a constant.


      #figure(figph[regions in the plane], alt: "regions in the plane", caption: none)


      In the polar plane, the coordinate grid lines are circles centered at the pole, with equations #math.equation(block: false, alt: "r equals k")[$r = k$], and lines through the pole, with equations #math.equation(block: false, alt: "θ equals k")[$θ = k$].


      #examplebox("Example 3")[][
Sketch the region designated by each set of inequalities.


          + #math.equation(block: false, alt: "1 less than or equal to r less than 3")[$1 ≤ r < 3$]
+ #math.equation(block: false, alt: "0 less than or equal to r less than or equal to 2 , the fraction π over 6 less than θ less than the fraction π over 3")[$0 ≤ r ≤ 2 , " " display(frac(π, 6)) < θ < display(frac(π, 3))$]


        
        #solutionbox[
+ The region consists of all points between 1 (inclusive) and 3 units from the pole. No restriction is given on #math.equation(block: false, alt: "θ")[$θ$], so the region is the annular ring shown in figure (a).#figure(figph[regions in polar coordinates], alt: "regions in polar coordinates", caption: none)
+ The region includes all points up to 2 units from the pole, but only between the lines #math.equation(block: false, alt: "θ equals the fraction π over 6")[$θ = display(frac(π, 6))$] and  #math.equation(block: false, alt: "θ equals the fraction π over 3")[$θ = display(frac(π, 3))$]. These curves bound the pie-shaped wedge (circular sector) shown in figure (b).
]
]


      
        Write inequalities to describe each region shown below.


        #figure(figph[regions], alt: "regions", caption: none)


        
          + #math.equation(block: false, alt: "r greater than or equal to 2 , the fraction 3 π over 4 less than θ less than or equal to the fraction 5 π over 4")[$r ≥ 2 , " " display(frac(3 π, 4)) < θ ≤ display(frac(5 π, 4))$]
+ #math.equation(block: false, alt: "2 less than r less than 3")[$2 < r < 3$], #math.equation(block: false, alt: "0 less than θ less than π")[$" " 0 < θ < π$]


        
      
    
    
      
=== Converting Between Polar and Cartesian Coordinates


      Using trigonometry, it is easy to convert from polar coordinates to Cartesian, or vice versa. The figure below shows how the two sets of coordinates are related.


      If we know the polar coordinates #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$], we can compute the Cartesian coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] as the legs of a right triangle.


      #math.equation(block: true, alt: "the fraction x over r equals cos ⁡ open parenthesis θ close parenthesis , so x equals r cos ⁡ open parenthesis θ close parenthesis; the fraction y over r equals sin ⁡ open parenthesis θ close parenthesis , so y equals r sin ⁡ open parenthesis θ close parenthesis")[$display(frac(x, r)) = cos ( θ ) , " " " " " " " " "so" " " " " " " " " x = r cos ( θ ) \ display(frac(y, r)) = sin ( θ ) , " " " " " " " " "so" " " " " " " " " y = r sin ( θ )$]
      #figure(figph[rectangular and polar coordinates], alt: "rectangular and polar coordinates", caption: none)


      #examplebox("Example 4")[][
Find the Cartesian coordinates of the point with polar coordinates #math.equation(block: false, alt: "open parenthesis the square root of 2 , the fraction 3 π over 4 close parenthesis")[$( sqrt(2) , display(frac(3 π, 4)) )$].


        
        #solutionbox[
The polar coordinates tell us that #math.equation(block: false, alt: "r equals the square root of 2")[$r = sqrt(2)$] and #math.equation(block: false, alt: "θ equals the fraction 3 π over 4")[$θ = display(frac(3 π, 4))$], as shown below.


          The #math.equation(block: false, alt: "x")[$x$]-coordinate of the point is


          #math.equation(block: true, alt: "x, equals r cos ⁡ open parenthesis θ close parenthesis equals the square root of 2 cos ⁡ open parenthesis the fraction 3 π over 4 close parenthesis; equals the square root of 2 times the fraction minus 1 over the square root of 2 equals minus 1")[$x & = r cos ( θ ) = sqrt(2) cos ( display(frac(3 π, 4)) ) \  & = sqrt(2) ⋅ display(frac(− 1, sqrt(2))) = − 1$]
          and its #math.equation(block: false, alt: "y")[$y$]-coordinate is


          #figure(figph[point], alt: "point", caption: none)


          #math.equation(block: true, alt: "y equals r sin ⁡ open parenthesis θ close parenthesis equals the square root of 2 sin ⁡ open parenthesis the fraction 3 π over 4 close parenthesis equals the square root of 2 times the fraction 1 over the square root of 2 equals 1")[$y = r sin ( θ ) = sqrt(2) sin ( display(frac(3 π, 4)) ) = sqrt(2) ⋅ display(frac(1, sqrt(2))) = 1$]
          Thus, the Cartesian coordinates are #math.equation(block: false, alt: "open parenthesis minus 1 , 1 close parenthesis")[$( − 1 , 1 )$].
]
]


      
        Find the Cartesian coordinates of the point with polar coordinates #math.equation(block: false, alt: "open parenthesis 4 , the fraction 5 π over 3 close parenthesis")[$( 4 , display(frac(5 π, 3)) )$].


        
          #math.equation(block: true, alt: "open parenthesis 2 , minus 2 the square root of 3 close parenthesis")[$( 2 , − 2 sqrt(3) )$]


        
      
      If we know the Cartesian coordinates of a point, we can compute the polar coordinates as follows.


      #math.equation(block: true, alt: "x squared plus y squared equals r squared , so r equals the square root of x squared plus y squared")[$x^(2) + y^(2) = r^(2) , " " " " "so" " " " " r = sqrt(x^(2) + y^(2))$]
      #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$tan ( θ ) = display(frac(y, x))$]
      To find the angle #math.equation(block: false, alt: "θ")[$θ$], it is not enough to compute #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the fraction y over x close parenthesis")[$tan^(− 1) ( display(frac(y, x)) )$]; we must choose the angle in the same quadrant as the given point.


      #examplebox("Example 5")[][
Find the polar coordinates of the point with Cartesian coordinates #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 , minus the fraction the square root of 3 over 2 close parenthesis")[$( − display(frac(1, 2)) , − display(frac(sqrt(3), 2)) )$].


        
        #solutionbox[
We first compute the value of #math.equation(block: false, alt: "r")[$r$].


          #math.equation(block: true, alt: "r squared equals x squared plus y squared equals open parenthesis minus the fraction 1 over 2 close parenthesis squared plus open parenthesis minus the fraction the square root of 3 over 2 close parenthesis squared equals the fraction 1 over 4 plus the fraction 3 over 4 equals 1")[$r^(2) = x^(2) + y^(2) = attach(( − display(frac(1, 2)) ), t: 2) + attach(( − display(frac(sqrt(3), 2)) ), t: 2) = display(frac(1, 4)) + display(frac(3, 4)) = 1$]
          Thus, #math.equation(block: false, alt: "r equals the square root of 1 equals 1")[$r = sqrt(1) = 1$], as shown at right. To find , we compute


          #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x equals the fraction minus the fraction the square root of 3 over 2 over minus the fraction 1 over 2 equals the square root of 3")[$tan ( θ ) = display(frac(y, x)) = display(frac(− display(frac(sqrt(3), 2)), − display(frac(1, 2)))) = sqrt(3)$]
          There are infinitely many values of #math.equation(block: false, alt: "θ")[$θ$] that satisfy this equation, but our point lies in the third quadrant, so we take #math.equation(block: false, alt: "θ equals the fraction 4 π over 3")[$θ = display(frac(4 π, 3))$].


          #figure(figph[polar], alt: "polar", caption: none)


          You can check that the polar coordinates #math.equation(block: false, alt: "open parenthesis 1 , the fraction 4 π over 3 close parenthesis")[$( 1 , display(frac(4 π, 3)) )$] convert back to Cartesian coordinates #math.equation(block: false, alt: "open parenthesis minus the fraction 1 over 2 , minus the fraction the square root of 3 over 2 close parenthesis")[$( − display(frac(1, 2)) , − display(frac(sqrt(3), 2)) )$].
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In the previous Example, the angle #math.equation(block: false, alt: "θ")[$θ$] is not given by #math.equation(block: false, alt: "tan to the power minus 1 ⁡ open parenthesis the square root of 3 close parenthesis equals the fraction π over 3")[$tan^(− 1) ( sqrt(3) ) = display(frac(π, 3))$], because the given point lies in the third quadrant. Remember that there are always two angles between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "2 π")[$2 π$] with a given tangent.
]


      
        Find polar coordinates for the points with Cartesian coordinates #math.equation(block: false, alt: "open parenthesis 2 , minus 5 close parenthesis")[$( 2 , − 5 )$]. Give decimal approximations rounded to hundredths.


        
          #math.equation(block: true, alt: "open parenthesis 5.39 , 5.09 close parenthesis")[$( 5.39 , 5.09 )$]


        
      
      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Conversion Equations]


        + To convert from polar coordinates #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] to Cartesian: 
        #math.equation(block: true, alt: "x, equals r cos ⁡ open parenthesis θ close parenthesis; y, equals r sin ⁡ open parenthesis θ close parenthesis")[$bold(italic(x)) & " " = bold(italic(r)) bold(cos) ( bold(italic(θ)) ) \ bold(italic(y)) & " " = bold(italic(r)) bold(sin) ( bold(italic(θ)) )$]
+ To convert from Cartesian coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] to polar:
        #math.equation(block: true, alt: "r equals the square root of x squared plus y squared; tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$& bold(italic(r)) = sqrt(bold(italic(x))^(2) + bold(italic(y))^(2)) \  & bold(tan) ( bold(italic(θ)) ) = display(frac(bold(italic(y)), bold(italic(x))))$] #linebreak() where the choice of #math.equation(block: false, alt: "θ")[$θ$] depends on the quadrant.
]


    
    
      
=== Equations in Polar Coordinates


      Polar coordinates are useful for studying objects or phenomena that have radial symmetry, such as circles, spheres, and cylinders, or the central forces (those that act equally in all directions), such as gravity and electric charge. The equations describing these objects are often simpler in polar coordinates than they are in Cartesian coordinates.


      It is not hard to convert an equation from Cartesian to polar coordinates:  we replace each #math.equation(block: false, alt: "x")[$x$] with #math.equation(block: false, alt: "r cos ⁡ open parenthesis θ close parenthesis")[$r cos ( θ )$] and each #math.equation(block: false, alt: "y")[$y$] with #math.equation(block: false, alt: "r sin ⁡ open parenthesis θ close parenthesis")[$r sin ( θ )$].


      #examplebox("Example 6")[][
Convert the equation #math.equation(block: false, alt: "2 x plus 3 y equals 6")[$2 x + 3 y = 6$] to polar coordinates.


        
        #solutionbox[
Replacing by #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "r cos ⁡ open parenthesis θ close parenthesis")[$r cos ( θ )$] and #math.equation(block: false, alt: "y")[$y$] by #math.equation(block: false, alt: "r sin ⁡ open parenthesis θ close parenthesis")[$r sin ( θ )$], we get


          #math.equation(block: true, alt: "2 r cos ⁡ open parenthesis θ close parenthesis plus 3 r sin ⁡ open parenthesis θ close parenthesis equals 6")[$2 r cos ( θ ) + 3 r sin ( θ ) = 6$]
          If we solve for #math.equation(block: false, alt: "r")[$r$] in terms of #math.equation(block: false, alt: "θ")[$θ$], we find


          #math.equation(block: true, alt: "r open parenthesis 2 cos ⁡ open parenthesis θ close parenthesis plus 3 sin ⁡ open parenthesis θ close parenthesis close parenthesis equals 6; r equals the fraction 6 over 2 cos ⁡ open parenthesis θ close parenthesis plus 3 sin ⁡ open parenthesis θ close parenthesis")[$& r ( 2 cos ( θ ) + 3 sin ( θ ) ) = 6 \  & r = display(frac(6, 2 cos ( θ ) + 3 sin ( θ )))$]
          Neither of these equations is simpler than the Cartesian version, but that should not surprise us, because the original equation describes a line, and Cartesian coordinates are better suited to rectilinear graphs.
]
]


      
        Convert the equation #math.equation(block: false, alt: "x squared plus y squared equals 4")[$x^(2) + y^(2) = 4$] to polar coordinates.


        
          Note that


          #math.equation(block: true, alt: "x squared plus y squared, equals r squared cos squared ⁡ open parenthesis θ close parenthesis plus r squared sin squared ⁡ open parenthesis θ close parenthesis; equals r squared open parenthesis cos squared ⁡ open parenthesis θ close parenthesis plus sin squared ⁡ open parenthesis θ close parenthesis close parenthesis equals r squared")[$x^(2) + y^(2) & = r^(2) cos^(2) ( θ ) + r^(2) sin^(2) ( θ ) \  & = r^(2) ( cos^(2) ( θ ) + sin^(2) ( θ ) ) = r^(2)$]
          so we have #math.equation(block: false, alt: "r squared equals 4")[$r^(2) = 4$], or #math.equation(block: false, alt: "r equals 2")[$r = 2$], a circle of raidus 2.


        
      
      Perhaps you recognize the equation in the previous exercise as a circle. Note also that its polar equation has a simpler form than the familiar Cartesian version.


      Converting equations from polar to Cartesian form is not as straightforward as the other direction (and is sometimes impossible).  Here are some strategies to try:


      - Expressions of the form #math.equation(block: false, alt: "r cos ⁡ open parenthesis θ close parenthesis")[$r cos ( θ )$]  or #math.equation(block: false, alt: "r sin ⁡ open parenthesis θ close parenthesis")[$r sin ( θ )$] can be replaced by #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$], respectively.
- #math.equation(block: false, alt: "r squared")[$r^(2)$] can be replaced by #math.equation(block: false, alt: "x squared plus y squared")[$x^(2) + y^(2)$].
- #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$] can be replaced by #math.equation(block: false, alt: "the fraction y over x")[$display(frac(y, x))$].


      #examplebox("Example 7")[][
Convert the equation #math.equation(block: false, alt: "r equals 3 cos ⁡ open parenthesis θ close parenthesis")[$r = 3 cos ( θ )$] to Cartesian coordinates.


        
        #solutionbox[
We multiply both sides of the equation by #math.equation(block: false, alt: "r")[$r$] to obtain


          #math.equation(block: true, alt: "r squared equals 3 r cos ⁡ open parenthesis θ close parenthesis")[$r^(2) = 3 r cos ( θ )$]
          Now we can replace the left side by #math.equation(block: false, alt: "x squared plus y squared")[$x^(2) + y^(2)$] and the right side by #math.equation(block: false, alt: "3 x")[$3 x$] to obtain


          #math.equation(block: true, alt: "x squared plus y squared equals 3 x")[$x^(2) + y^(2) = 3 x$]
          This is the equation of a circle. By completing the square in each variable, we can write the equation in standard form,


          #math.equation(block: true, alt: "open parenthesis x minus the fraction 3 over 2 close parenthesis squared plus y squared equals the fraction 9 over 4")[$attach(( x − display(frac(3, 2)) ), t: 2) + y^(2) = display(frac(9, 4))$]
          The center of the circle is #math.equation(block: false, alt: "open parenthesis the fraction 3 over 2 , 0 close parenthesis")[$( display(frac(3, 2)) , 0 )$], and its radius is #math.equation(block: false, alt: "the fraction 3 over 2")[$display(frac(3, 2))$].
]
]


      
        Convert the equation #math.equation(block: false, alt: "r equals tan ⁡ open parenthesis θ close parenthesis")[$r = tan ( θ )$] to Cartesian coordinates.


        
          Start by squaring both sides and then replace each side by its Cartesian form. Then solve for #math.equation(block: false, alt: "y squared")[$y^(2)$] to get #math.equation(block: false, alt: "y squared equals the fraction x to the power 4 over 1 minus x squared")[$y^(2) = display(frac(x^(4), 1 − x^(2)))$]


        
      
      Review the following skills you will need for this section.


      #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Recall the standard form for the equation of a circle of radius #math.equation(block: false, alt: "r")[$r$] centered at #math.equation(block: false, alt: "open parenthesis h , k close parenthesis")[$( h , k )$]:


        #math.equation(block: true, alt: "open parenthesis x minus h close parenthesis squared plus open parenthesis y minus k close parenthesis squared equals r squared")[$( x − h )^(2) + ( y − k )^(2) = r^(2)$]
        Use completing the square to write each equation in standard form.


        + #math.equation(block: false, alt: "x squared plus y squared minus 4 x plus 2 y minus 4 equals 0")[$x^(2) + y^(2) − 4 x + 2 y − 4 = 0$]
+ #math.equation(block: false, alt: "x squared plus y squared equals 88 plus 18 y")[$x^(2) + y^(2) = 88 + 18 y$]
+ #math.equation(block: false, alt: "x squared plus y squared plus 25 equals 9 plus 10 x plus 6 y")[$x^(2) + y^(2) + 25 = 9 + 10 x + 6 y$]
+ #math.equation(block: false, alt: "5 y squared equals 75 minus 5 x squared plus 10 x")[$5 y^(2) = 75 − 5 x^(2) + 10 x$]


        #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$]


        Algebra Refresher Answers


        + #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y plus 1 close parenthesis squared equals 3 squared")[$( x − 2 )^(2) + ( y + 1 )^(2) = 3^(2)$]
+ #math.equation(block: false, alt: "x squared plus open parenthesis y minus 9 close parenthesis squared equals 13 squared")[$x^(2) + ( y − 9 )^(2) = 13^(2)$]
+ #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis squared plus open parenthesis y minus 3 close parenthesis squared equals open parenthesis 3 the square root of 2 close parenthesis squared")[$( x − 5 )^(2) + ( y − 3 )^(2) = attach(( 3 sqrt(2) ), t: 2)$]
+ #math.equation(block: false, alt: "open parenthesis x minus 1 close parenthesis squared plus y squared equals 4 squared")[$( x − 1 )^(2) + y^(2) = 4^(2)$]
]


    
    
      
=== Section 10.1 Summary


      
        
==== Vocabulary


        - Pole
- Polar axis
- Polar coordinates


      
      
        
==== Concepts


        + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Polar Coordinates]

The #strong[polar coordinates] of a point #math.equation(block: false, alt: "P")[$P$] in the plane are #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$], where

- #math.equation(block: false, alt: "vertical bar r vertical bar")[$| r |$]  is the #strong[distance] from  #math.equation(block: false, alt: "P")[$P$]  to the pole,
- #math.equation(block: false, alt: "θ")[$θ$]   is the #strong[angle] measured counterclockwise from the polar axis to the ray through #math.equation(block: false, alt: "P")[$P$] from the pole.

#figure(figph[rectangular and polar coordinates], alt: "rectangular and polar coordinates", caption: none)
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Non-Uniqueness of Polar Coordinates]

+ Any point with polar coordinates #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] also has coordinates #math.equation(block: false, alt: "open parenthesis r , θ plus 2 k π close parenthesis")[$( r , θ + 2 k π )$], where #math.equation(block: false, alt: "k")[$k$] is an integer.
+ The point #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] can also be designated by #math.equation(block: false, alt: "open parenthesis minus r , θ plus π close parenthesis")[$( − r , θ + π )$].
+ The pole has coordinates #math.equation(block: false, alt: "open parenthesis 0 , θ close parenthesis")[$( 0 , θ )$], for any value of #math.equation(block: false, alt: "θ")[$θ$].
]
+ In the polar plane, the coordinate grid lines are circles centered at the pole, with equations #math.equation(block: false, alt: "r equals k")[$r = k$], and lines through the pole, with equations #math.equation(block: false, alt: "θ equals k")[$θ = k$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Conversion Equations]

+ To convert from polar coordinates #math.equation(block: false, alt: "open parenthesis r , θ close parenthesis")[$( r , θ )$] to Cartesian: 
            #math.equation(block: true, alt: "x, equals r cos ⁡ open parenthesis θ close parenthesis; y, equals r sin ⁡ open parenthesis θ close parenthesis")[$x & = r cos ( θ ) \ y & = r sin ( θ )$]
+ To convert from Cartesian coordinates #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] to polar:
            #math.equation(block: true, alt: "r equals the square root of x squared plus y squared; tan ⁡ open parenthesis θ close parenthesis equals the fraction y over x")[$& r = sqrt(x^(2) + y^(2)) \  & tan ( θ ) = display(frac(y, x))$]

where the choice of #math.equation(block: false, alt: "θ")[$θ$] depends on the quadrant.
]
+ To convert an equation from Cartesian to polar coordinates, we replace each #math.equation(block: false, alt: "x")[$x$] with #math.equation(block: false, alt: "r cos ⁡ open parenthesis θ close parenthesis")[$r cos ( θ )$] and each with #math.equation(block: false, alt: "y")[$y$] with #math.equation(block: false, alt: "r sin ⁡ open parenthesis θ close parenthesis")[$r sin ( θ )$]. To convert an equation from polar to Cartesian coordinates, look for expressions of the form #math.equation(block: false, alt: "r cos ⁡ open parenthesis θ close parenthesis , r sin ⁡ open parenthesis θ close parenthesis , r squared")[$r cos ( θ ) , " " r sin ( θ ) , " " r^(2)$], or #math.equation(block: false, alt: "tan ⁡ open parenthesis θ close parenthesis")[$tan ( θ )$].


      
      
        
====  Study Questions


        + Why is the Cartesian coordinate system sometimes called the rectangular system?
+ What do the graphs of the polar equations #math.equation(block: false, alt: "r equals k")[$r = k$] and #math.equation(block: false, alt: "θ equals k")[$θ = k$], where #math.equation(block: false, alt: "k")[$k$] is a constant, look like?
+ Fill in the blank:  the point with polar coordinates #math.equation(block: false, alt: "open parenthesis minus 6 , θ close parenthesis")[$( − 6 , θ )$] also has coordinates #math.equation(block: false, alt: "open parenthesis 6 , ? close parenthesis")[$( 6 , ? )$].
+ What does the region described by #math.equation(block: false, alt: "a less than or equal to r less than or equal to b")[$a ≤ r ≤ b$] look like?  What does the region described by #math.equation(block: false, alt: "a less than or equal to θ less than or equal to b")[$a ≤ θ ≤ b$] look like?
+ Sketch a figure that explains how to convert between polar and Cartesian coordinates.


      
      
        
==== Skills


        + Plot points in polar coordinates    \#1–8
+ Write polar coordinates for points    \#9–16
+ Convert Cartesian coordinates to polar    \#17–24
+ Convert Polar coordinates to Cartesian    \#25–32
+ Write alternate versions of polar coordinates    \#33–38
+ Sketch a region described by polar inequalities    \#39–44
+ Write polar inequalities to describe a region    \#45–50
+ Convert an equation to Cartesian coordinates    \#51–64
+ Convert an equation to polar coordinates    \#65–72


      
    
    
      
=== Homework 10-1


      For Problems 1–8, use the grid below to plot the points whose polar coordinates are given.


      #figure(figph[polar grid], alt: "polar grid", caption: none)


      
        #math.equation(block: true, alt: "open parenthesis 2 , the fraction 2 π over 3 close parenthesis")[$( 2 , display(frac(2 π, 3)) )$]


        
          #figure(figph[polar plot], alt: "polar plot", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis 3 , the fraction 5 π over 4 close parenthesis")[$( 3 , display(frac(5 π, 4)) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis 3 , the fraction 3 π over 2 close parenthesis")[$( 3 , display(frac(3 π, 2)) )$]


        
          #figure(figph[polar plot], alt: "polar plot", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis 4 , π close parenthesis")[$( 4 , π )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 4 , the fraction π over 6 close parenthesis")[$( − 4 , display(frac(π, 6)) )$]


        
          #figure(figph[polar plot], alt: "polar plot", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 1 , the fraction π over 2 close parenthesis")[$( − 1 , display(frac(π, 2)) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 1 , the fraction 7 π over 4 close parenthesis")[$( − 1 , display(frac(7 π, 4)) )$]


        
          #figure(figph[polar plot], alt: "polar plot", caption: none)


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 2 , the fraction 5 π over 3 close parenthesis")[$( − 2 , display(frac(5 π, 3)) )$]


      
      For Problems 9–16, give polar coordinates for each point shown below, with #math.equation(block: false, alt: "r greater than or equal to 0")[$r ≥ 0$] and #math.equation(block: false, alt: "0 less than or equal to θ less than 2 π")[$0 ≤ θ < 2 π$].


      #figure(figph[polar grid], alt: "polar grid", caption: none)


      
        #math.equation(block: true, alt: "A")[$A$]


        
          #math.equation(block: true, alt: "open parenthesis 5 , the fraction 3 π over 4 close parenthesis")[$( 5 , display(frac(3 π, 4)) )$]


        
      
      
        #math.equation(block: true, alt: "B")[$B$]


      
      
        #math.equation(block: true, alt: "C")[$C$]


        
          #math.equation(block: true, alt: "open parenthesis 1 , π close parenthesis")[$( 1 , π )$]


        
      
      
        #math.equation(block: true, alt: "D")[$D$]


      
      
        #math.equation(block: true, alt: "E")[$E$]


        
          #math.equation(block: true, alt: "open parenthesis 3 , the fraction 4 π over 3 close parenthesis")[$( 3 , display(frac(4 π, 3)) )$]


        
      
      
        #math.equation(block: true, alt: "F")[$F$]


      
      
        #math.equation(block: true, alt: "G")[$G$]


        
          #math.equation(block: true, alt: "open parenthesis 2 , the fraction π over 12 close parenthesis")[$( 2 , display(frac(π, 12)) )$]


        
      
      
        #math.equation(block: true, alt: "H")[$H$]


      
      For Problems 17–24, convert the polar coordinates to Cartesian coordinates.


      
        #math.equation(block: true, alt: "open parenthesis 6 , the fraction 2 π over 3 close parenthesis")[$( 6 , display(frac(2 π, 3)) )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 3 , 3 the square root of 3 close parenthesis")[$( − 3 , 3 sqrt(3) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 5 , the fraction 7 π over 6 close parenthesis")[$( 5 , display(frac(7 π, 6)) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 3 , the fraction 3 π over 4 close parenthesis")[$( − 3 , display(frac(3 π, 4)) )$]


        
          #math.equation(block: true, alt: "open parenthesis the fraction 3 over the square root of 2 , the fraction minus 3 over the square root of 2 close parenthesis")[$( display(frac(3, sqrt(2))) , display(frac(− 3, sqrt(2))) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 4 , the fraction 5 π over 3 close parenthesis")[$( − 4 , display(frac(5 π, 3)) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis 2.4 , 3.6 close parenthesis")[$( 2.4 , 3.6 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 2.15 , minus 1.06 close parenthesis")[$( − 2.15 , − 1.06 )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 1.7 , 5.2 close parenthesis")[$( 1.7 , 5.2 )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 2 , 1.5 close parenthesis")[$( − 2 , 1.5 )$]


        
          #math.equation(block: true, alt: "open parenthesis minus 0.14 , minus 1.99 close parenthesis")[$( − 0.14 , − 1.99 )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 3 , 4.5 close parenthesis")[$( − 3 , 4.5 )$]


      
      For Problems 25–32, convert the Cartesian coordinates to polar coordinates with #math.equation(block: false, alt: "r greater than or equal to 0")[$r ≥ 0$] and #math.equation(block: false, alt: "0 less than or equal to θ less than or equal to 2 π")[$0 ≤ θ ≤ 2 π$].  Give exact values for #math.equation(block: false, alt: "r")[$r$] and #math.equation(block: false, alt: "θ")[$θ$].


      
        #math.equation(block: true, alt: "open parenthesis 7 , 7 close parenthesis")[$( 7 , 7 )$]


        
          #math.equation(block: true, alt: "open parenthesis 7 the square root of 2 , the fraction π over 4 close parenthesis")[$( 7 sqrt(2) , display(frac(π, 4)) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus 7 , minus 7 close parenthesis")[$( − 7 , − 7 )$]


      
      
        #math.equation(block: true, alt: "open parenthesis the square root of 6 , minus the square root of 2 close parenthesis")[$( sqrt(6) , − sqrt(2) )$]


        
          #math.equation(block: true, alt: "open parenthesis 2 the square root of 2 , the fraction 11 π over 6 close parenthesis")[$( 2 sqrt(2) , display(frac(11 π, 6)) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis minus the square root of 3 , 3 close parenthesis")[$( − sqrt(3) , 3 )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 3 , minus 2 close parenthesis")[$( − 3 , − 2 )$]


        
          #math.equation(block: true, alt: "open parenthesis the square root of 13 , π plus tan to the power minus 1 ⁡ the fraction 2 over 3 close parenthesis")[$( sqrt(13) , π + tan^(− 1) display(frac(2, 3)) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$]


      
      
        #math.equation(block: true, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$]


        
          #math.equation(block: true, alt: "open parenthesis 2 , π close parenthesis")[$( 2 , π )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 0 , minus 5 close parenthesis")[$( 0 , − 5 )$]


      
      In Problems 33–38, polar coordinates are given.


      + Convert to polar coordinates with #math.equation(block: false, alt: "r less than or equal to 0 , 0 less than or equal to θ less than 2 π")[$r ≤ 0 , " " 0 ≤ θ < 2 π$].
+ Convert to polar coordinates with #math.equation(block: false, alt: "r greater than or equal to 0 , minus 2 π less than θ less than or equal to 0")[$r ≥ 0 , " " − 2 π < θ ≤ 0$].


      
        #math.equation(block: true, alt: "open parenthesis 2 , the fraction 5 π over 6 close parenthesis")[$( 2 , display(frac(5 π, 6)) )$]


        
          + #math.equation(block: false, alt: "open parenthesis minus 2 , the fraction 11 π over 6 close parenthesis")[$( − 2 , display(frac(11 π, 6)) )$]
+ #math.equation(block: false, alt: "open parenthesis 2 , the fraction minus 7 π over 6 close parenthesis")[$( 2 , display(frac(− 7 π, 6)) )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 1 , the fraction 4 π over 3 close parenthesis")[$( 1 , display(frac(4 π, 3)) )$]


      
      
        #math.equation(block: true, alt: "open parenthesis 3 , π close parenthesis")[$( 3 , π )$]


        
          + #math.equation(block: false, alt: "open parenthesis minus 3 , 0 close parenthesis")[$( − 3 , 0 )$]
+ #math.equation(block: false, alt: "open parenthesis 3 , minus π close parenthesis")[$( 3 , − π )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]


      
      
        #math.equation(block: true, alt: "open parenthesis 2.3 , 5.2 close parenthesis")[$( 2.3 , 5.2 )$]


        
          + #math.equation(block: false, alt: "open parenthesis minus 2.3 , 2.06 close parenthesis")[$( − 2.3 , 2.06 )$]
+ #math.equation(block: false, alt: "open parenthesis 2.3 , minus 1.08 close parenthesis")[$( 2.3 , − 1.08 )$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 1.2 , 1.2 close parenthesis")[$( 1.2 , 1.2 )$]


      
      For Problems 39–44, sketch the region described by the inequalities.


      
        #math.equation(block: true, alt: "r less than or equal to 3")[$r ≤ 3$]


        
          #figure(figph[polar plot], alt: "polar plot", caption: none)


        
      
      
        #math.equation(block: true, alt: "2 less than or equal to r less than or equal to 3")[$2 ≤ r ≤ 3$]


      
      
        #math.equation(block: true, alt: "the fraction π over 6 less than or equal to θ less than or equal to the fraction π over 3")[$display(frac(π, 6)) ≤ θ ≤ display(frac(π, 3))$]


        
          #figure(figph[polar grid], alt: "polar grid", caption: none)


        
      
      
        #math.equation(block: true, alt: "0 less than or equal to θ less than or equal to the fraction π over 2")[$0 ≤ θ ≤ display(frac(π, 2))$]


      
      
        #math.equation(block: true, alt: "r greater than or equal to 4 , the fraction π over 2 less than or equal to θ less than or equal to the fraction 3 π over 4")[$r ≥ 4 , " " display(frac(π, 2)) ≤ θ ≤ display(frac(3 π, 4))$]


        
          #figure(figph[polar grid], alt: "polar grid", caption: none)


        
      
      
        #math.equation(block: true, alt: "2 less than or equal to r less than or equal to 3 , the fraction 5 π over 3 less than or equal to θ less than or equal to the fraction 11 π over 6")[$2 ≤ r ≤ 3 , " " display(frac(5 π, 3)) ≤ θ ≤ display(frac(11 π, 6))$]


      
      For Problems 45–50, write inequalities to describe the region.


      
        #figure(figph[polar region], alt: "polar region", caption: none)


        
          #math.equation(block: true, alt: "r greater than or equal to 0 , the fraction π over 6 less than or equal to θ less than or equal to the fraction π over 3")[$r ≥ 0 , " " display(frac(π, 6)) ≤ θ ≤ display(frac(π, 3))$]


        
      
      
        #figure(figph[polar grid], alt: "polar grid", caption: none)


      
      
        #figure(figph[polar region], alt: "polar region", caption: none)


        
          #math.equation(block: true, alt: "r greater than or equal to 1 , the fraction π over 2 less than or equal to θ less than or equal to π")[$r ≥ 1 , " " display(frac(π, 2)) ≤ θ ≤ π$]


        
      
      
        #figure(figph[polar region], alt: "polar region", caption: none)


      
      
        #figure(figph[polar region], alt: "polar region", caption: none)


        
          #math.equation(block: true, alt: "minus 1 less than or equal to r less than or equal to 1 , the fraction 3 π over 4 less than or equal to θ less than or equal to π")[$− 1 ≤ r ≤ 1 , " " display(frac(3 π, 4)) ≤ θ ≤ π$]


        
      
      
        #figure(figph[polar region], alt: "polar region", caption: none)


      
      For Problems 51–64, convert the equation into Cartesian coordinates.


      
        #math.equation(block: true, alt: "r squared equals 2")[$r^(2) = 2$]


        
          #math.equation(block: true, alt: "x squared plus y squared equals 2")[$x^(2) + y^(2) = 2$]


        
      
      
        #math.equation(block: true, alt: "r equals 2")[$r = 2$]


      
      
        #math.equation(block: true, alt: "r equals 4 cos ⁡ open parenthesis θ close parenthesis")[$r = 4 cos ( θ )$]


        
          #math.equation(block: true, alt: "x squared plus y squared equals 4 x")[$x^(2) + y^(2) = 4 x$]


        
      
      
        #math.equation(block: true, alt: "r equals sin ⁡ open parenthesis θ close parenthesis")[$r = sin ( θ )$]


      
      
        #math.equation(block: true, alt: "r equals csc ⁡ open parenthesis θ close parenthesis")[$r = csc ( θ )$]


        
          #math.equation(block: true, alt: "y equals 1")[$y = 1$]


        
      
      
        #math.equation(block: true, alt: "r equals 2 sec ⁡ open parenthesis θ close parenthesis")[$r = 2 sec ( θ )$]


      
      
        #math.equation(block: true, alt: "tan ⁡ open parenthesis θ close parenthesis equals 2")[$tan ( θ ) = 2$]


        
          #math.equation(block: true, alt: "y equals 2 x")[$y = 2 x$]


        
      
      
        #math.equation(block: true, alt: "r squared equals tan ⁡ open parenthesis θ close parenthesis")[$r^(2) = tan ( θ )$]


      
      
        #math.equation(block: true, alt: "r sec ⁡ open parenthesis θ close parenthesis equals 3")[$r sec ( θ ) = 3$]


        
          #math.equation(block: true, alt: "x squared plus y squared equals 3 x")[$x^(2) + y^(2) = 3 x$]


        
      
      
        #math.equation(block: true, alt: "θ equals the fraction π over 3")[$θ = display(frac(π, 3))$]


      
      
        #math.equation(block: true, alt: "r equals the fraction 2 over 1 plus sin ⁡ open parenthesis θ close parenthesis")[$r = display(frac(2, 1 + sin ( θ )))$]


        
          #math.equation(block: true, alt: "x squared equals 4 minus 4 y")[$x^(2) = 4 − 4 y$]


        
      
      
        #math.equation(block: true, alt: "r equals the fraction 4 over 4 minus cos ⁡ open parenthesis θ close parenthesis")[$r = display(frac(4, 4 − cos ( θ )))$]


      
      
        #math.equation(block: true, alt: "2 r cos ⁡ open parenthesis θ close parenthesis plus r sin ⁡ open parenthesis θ close parenthesis equals 1")[$2 r cos ( θ ) + r sin ( θ ) = 1$]


        
          #math.equation(block: true, alt: "2 x plus y equals 1")[$2 x + y = 1$]


        
      
      
        #math.equation(block: true, alt: "tan ⁡ θ equals r cos ⁡ θ minus 2")[$tan θ = r cos θ − 2$]


      
      For Problems 65–72, convert the equation into polar coordinates.


      
        #math.equation(block: true, alt: "x equals 2")[$x = 2$]


        
          #math.equation(block: true, alt: "r equals 2 sec ⁡ open parenthesis θ close parenthesis")[$r = 2 sec ( θ )$]


        
      
      
        #math.equation(block: true, alt: "y squared equals 2 x minus x squared")[$y^(2) = 2 x − x^(2)$]


      
      
        #math.equation(block: true, alt: "2 x y equals 1")[$2 x y = 1$]


        
          #math.equation(block: true, alt: "2 r squared equals sec ⁡ open parenthesis θ close parenthesis csc ⁡ open parenthesis θ close parenthesis")[$2 r^(2) = sec ( θ ) csc ( θ )$]


        
      
      
        #math.equation(block: true, alt: "y equals minus x")[$y = − x$]


      
      
        #math.equation(block: true, alt: "y squared equals 4 x")[$y^(2) = 4 x$]


        
          #math.equation(block: true, alt: "r equals 4 cot ⁡ open parenthesis θ close parenthesis csc ⁡ open parenthesis θ close parenthesis")[$r = 4 cot ( θ ) csc ( θ )$]


        
      
      
        #math.equation(block: true, alt: "x squared minus y squared equals 1")[$x^(2) − y^(2) = 1$]


      
      
        #math.equation(block: true, alt: "x squared plus y squared equals 4 the square root of x squared plus y squared")[$x^(2) + y^(2) = 4 sqrt(x^(2) + y^(2))$]


        
          #math.equation(block: true, alt: "r equals 4")[$r = 4$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis x squared plus y squared close parenthesis cubed equals 9 x squared y squared")[$( x^(2) + y^(2) )^(3) = 9 x^(2) y^(2)$]


      
      
        Use the Law of Cosines to prove the distance formula in polar coordinates:


        #math.equation(block: true, alt: "d equals the square root of r sub 1 squared plus r sub 2 squared minus 2 r sub 1 r sub 2 cos ⁡ open parenthesis θ sub 2 minus θ sub 1 close parenthesis")[$d = sqrt(r_(1)^(2) + r_(2)^(2) − 2 r_(1) r_(2) cos ( θ_(2) − θ_(1) ))$]
        
          #math.equation(block: true, alt: "d, equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared; equals the square root of open parenthesis r sub 2 cos ⁡ open parenthesis θ sub 2 close parenthesis minus r sub 1 cos ⁡ open parenthesis θ sub 1 close parenthesis close parenthesis squared plus open parenthesis r sub 2 sin ⁡ open parenthesis θ sub 2 close parenthesis minus r sub 1 sin ⁡ open parenthesis θ sub 1 close parenthesis close parenthesis squared; equals the square root of r sub 2 squared cos squared ⁡ open parenthesis θ sub 2 close parenthesis minus 2 r sub 2 r sub 1 cos ⁡ open parenthesis θ sub 2 close parenthesis cos ⁡ open parenthesis θ sub 1 close parenthesis plus r sub 1 squared cos squared ⁡ open parenthesis θ su")[$d & = sqrt(( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2)) \  & = sqrt(( r_(2) cos ( θ_(2) ) − r_(1) cos ( θ_(1) ) )^(2) + ( r_(2) sin ( θ_(2) ) − r_(1) sin ( θ_(1) ) )^(2)) \  & = sqrt(r_(2)^(2) cos^(2) ( θ_(2) ) − 2 r_(2) r_(1) cos ( θ_(2) ) cos ( θ_(1) ) + r_(1)^(2) cos^(2) ( θ_(1) ) + r_(2)^(2) sin^(2) ( θ_(2) ) − 2 r_(2) r_(1) sin ( θ_(2) ) sin ( θ_(1) ) + r_(1)^(2) sin^(2) ( θ_(1) )) \  & = sqrt(r_(2)^(2) + r_(1)^(2) − 2 r_(2) r_(1) ( cos ( θ_(2) ) cos ( θ_(1) ) − sin ( θ_(2) ) sin ( θ_(1) ) )) \  & = sqrt(r_(1)^(2) + r_(2)^(2) − 2 r_(1) r_(2) cos ( θ_(2) − θ_(1) ))$]
        
      
      
        Show that the graph of #math.equation(block: false, alt: "r equals a cos ⁡ θ plus b sin ⁡ θ")[$r = a cos θ + b sin θ$] is a circle. Find its center and radius.